Consider a body of mass $m$ kg on a smooth inclined plane inclined at an angle of $\theta$ to the horizontal. Let $R$ be the magnitude of the normal reaction force acting on the body.

If I resolve forces vertically, I get $Rcos\theta=mg$

And if I resolve perpendicular to the plane, I get $R=mgcos\theta$

Clearly this is not possible. What am I doing wrong exactly?


closed as unclear what you're asking by Bill N, ZeroTheHero, Kyle Kanos, Sebastian Riese, knzhou Apr 25 '18 at 10:13

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  • $\begingroup$ What is $R$? A vector magnitude? What is $\theta$? A direction angle? With respect to what? $\endgroup$ – Bill N Apr 24 '18 at 20:17
  • $\begingroup$ @BillN I’ve edited the question. Hope it’s clearer now. $\endgroup$ – s.xw Apr 24 '18 at 20:20
  • $\begingroup$ I'm not sure if this is a duplicate, exactly, but it's closely related and might be a duplicate: physics.stackexchange.com/questions/176200/… $\endgroup$ – David Z Apr 24 '18 at 20:23
  • $\begingroup$ @npojo That should be an answer, not a comment $\endgroup$ – David Z Apr 24 '18 at 20:24
  • 1
    $\begingroup$ Are you using Newton's 1st law on an accelerating object? $\endgroup$ – Steeven Apr 24 '18 at 20:27

If the body is moving down the slope, than there is vertical acceleration and you cannot equate vertical forced. If the body is static, there must be an additional force such as friction which adds to the vertical direction

  • $\begingroup$ What if the body is going around a banked corner (circular motion), banked at angle $\theta$ to the horizontal? $\endgroup$ – s.xw Apr 24 '18 at 20:28
  • $\begingroup$ There is no way around this. If nothing opposes $mgsin\theta$ along the plane, the body will slip down. $\endgroup$ – npojo Apr 24 '18 at 20:38
  • $\begingroup$ But surely at the right speed, a body can negotiate a frictionless banked corner without slipping? $\endgroup$ – s.xw Apr 24 '18 at 20:43
  • $\begingroup$ It's not a question of negotiations but a meticulous sketch of the forces involved. $\endgroup$ – npojo Apr 24 '18 at 20:50

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